Solve Using the Quadratic Formula p^2-16p+13=-3p^2

$p^{2} - 16 p + 13 = - 3 p^{2}$

Step 1

Move all terms to the left side of the equation and simplify.

Add $3 p^{2}$ to both sides of the equation.

$p^{2} - 16 p + 13 + 3 p^{2} = 0$

Add $p^{2}$ and $3 p^{2}$ .

$4 p^{2} - 16 p + 13 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 4$ , $b = - 16$ , and $c = 13$ into the quadratic formula and solve for $p$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (4 \cdot 13)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 16$ to the power of $2$ .

$p = \frac{16 \pm \sqrt{256 - 4 \cdot 4 \cdot 13}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 13$ .

Multiply $- 4$ by $4$ .

$p = \frac{16 \pm \sqrt{256 - 16 \cdot 13}}{2 \cdot 4}$

Multiply $- 16$ by $13$ .

$p = \frac{16 \pm \sqrt{256 - 208}}{2 \cdot 4}$

Subtract $208$ from $256$ .

$p = \frac{16 \pm \sqrt{48}}{2 \cdot 4}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$p = \frac{16 \pm \sqrt{16 (3)}}{2 \cdot 4}$

Rewrite $16$ as $4^{2}$ .

$p = \frac{16 \pm \sqrt{4^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$p = \frac{16 \pm 4 \sqrt{3}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$p = \frac{16 \pm 4 \sqrt{3}}{8}$

Simplify $\frac{16 \pm 4 \sqrt{3}}{8}$ .

$p = \frac{4 \pm \sqrt{3}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$p = \frac{4 \pm \sqrt{3}}{2}$

Decimal Form:

$p = 2.86602540 \dots, 1.13397459 \dots$